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Def of rational map

Let X and Y be varieties. A rational map $\phi:X \to Y$ is an equivalence class of pairs $(U,\phi_U)$ where $U$ is a nonempty open subset of X, $\phi_U$ is a mophism of U to Y and where $(U,\phi_U)$ and $(V,\phi_V)$ are equivalent if $\phi_U$ and $\phi_V$ agree on $U \cap V$.

Question: Why a rational map is not in general a map of the set $X$ to $Y$.

Ref: Robin Hartshorne p24

I think the reason why a rational map is not a map as sets is that it dose not agree along the closed curve. image

Jean Billie
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    Well, I think your very description says it all: it is not a map, it is an equivalence class... – DonAntonio Oct 26 '19 at 11:01
  • Why? For any point $p \in X$, I think we can define a map by $\phi_U(p)$ for any $U$ including the point $p$. – Jean Billie Oct 26 '19 at 11:06
  • Yes, so what? That is a map...but a rational map as defined by yourself is an equivalence class ! – DonAntonio Oct 26 '19 at 11:07
  • For any rational map which is not a map, can we define a map by the previous comment? – Jean Billie Oct 26 '19 at 11:09
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    No. Have you looked at any examples of this? Things will become a lot clearer if you do - a good first example to start with would be the rational map of the affine line to itself given by inversion on the complement of zero. – KReiser Oct 26 '19 at 11:58
  • No, I have not ever seen any example of rational map. Your example mean $k \to k:x \mapsto -x$? – Jean Billie Oct 26 '19 at 12:09
  • No, I mean $x\mapsto \frac{1}{x}$. – KReiser Oct 26 '19 at 17:40

1 Answers1

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First in general, we denote a rational map with a dashed arrow $\phi : X \dashrightarrow Y$.

It's an equivalence class or maps, so what does it mean : it means that there EXISTS $(U,\phi_U)$ such that your rational map is the equivalence class of this element, not that for all $U$ there is a $\phi_U$.

Imagine $U=\mathbb{C}\subseteq \mathbb{P}^1$ and $\phi_U(x)=x^2$. Then your function has a pole at infinity so it cannot be extended further : there is no $(V,\phi_V)$ with $\infty \in V$ and $\phi_U=\phi_V$ on $U \cap V$, so there is no such $(V,\phi_V)$ in the equivalence class of $(U,\phi_U)$ !

Carot
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  • If $Y=\mathbb{P}$, then the rational map is a map of sets. However if $Y$ is an affine line, then the rational map is not a map of sets. Is my interpretation of your example correct? – Jean Billie Oct 26 '19 at 12:16
  • There is a "not" to add somewhere, but yes. – Carot Oct 26 '19 at 12:20
  • is rational map not a map in general in case that $X$ and $Y$ are projective varieties ? – Jean Billie Oct 26 '19 at 12:26
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    This rational map in your post actually does extend to an honest map $\Bbb P^1\to \Bbb P^1$. In homogeneous coordinates, it can be described as $[x:y]\mapsto [x^2:y^2]$. In general, for a rational map to $\Bbb P^n$, the indeterminacy locus of the map is codimension two in the source, which means that it turns in to an honest map if the source is dimension one or less like here. There are still rational maps of projective varieties which are not extendable to honest maps: see $\Bbb P^{n+1}\to\Bbb P^n$ by sending $[x_0:\cdots:x_{n+1}]\mapsto[x_0:\cdots:x_n]$ for $n>0$. – KReiser Oct 26 '19 at 17:56
  • Indeed. The idea here was to give a very simple example, because the confusion seemed to be a very basic one. – Carot Oct 27 '19 at 15:30
  • The map from $\mathbb{P}^{n+1}$ cannot send the element $[x_0: \cdots :x_{n+1}] = [0:\cdots :0:1]$ to the target space $\mathbb{P}^{n}$, because it will be $[0:\cdots :0]$ which is not an element but it is a rational map. This example is very simple and naive. I can understand. – Jean Billie Oct 28 '19 at 10:53